The key idea is to rephrase amicable pairs in terms of more familiar concepts. Let $N^k(v)$ denote the $k$th neighbor set of $v$, and $S$ the set of vertices of nonzero degree. Note that $N^2(v)$ contains $v$ for all $v \in S$, so a simple double-counting argument yields
$$\begin{align*} 2(\text{number of unordered amicable pairs}) + |S| &= \sum_{v \in S} |N^2(v)| \\ &= \sum_{v \in S} \left| \bigcup_{u \in N(v)} N(u) \right| \\ &\ge \sum_{v \in S} \max_{u \in N(v)} |N(u)| \\ &\ge \sum_{v \in S} \frac{1}{|N(v)|} \sum_{u \in N(v)} |N(u)| \quad (\text{by averaging over all } u \in N(v)) \\ &= \sum_{\text{edges } uv \in E(G)} \left( \frac{|N(u)|}{|N(v)|} + \frac{|N(v)|}{|N(u)|} \right) \\ &\ge 2|E(G)|, \end{align*}$$
where $E(G)$ denote the set of edges in $G$, and we use AM-GM in the last step.
Thus the number of (unordered) amicable pairs is at least $|E(G)| - |S|/2 \ge n^2/4 - n/2 = 2^{\binom{n/2}{2}}$, as desired.